Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm studying for a test, and I need help with this problem. I am not sure how to prove that this is not linear due to the notation. The comma is throwing me off.

Show that the transformation $T$ defined by $T(x_1,x_2)$ = $(x_1^2-2x_2, x_1+5x_2)$ is not linear.

I know that the definition of a linear transformation involves:

  1. $T(u+v)=T(u)+T(v)$ for all $u, v$ in the domain of $T$.

  2. $T(c*u) = c*T(u)$ for all scalars $c$ and all $u$ in the domain of $T$.

  3. $T(cu + dv) = cT(u) + dT(v)$ for all vectors $u, v$ in the domain of $T$ and all scalars $c, d$

  4. $T(0) = 0$ if $T$ is linear

However, I'm not sure how to use this definition with the specific function given.

Would $T(x_1 + x_2)$ = $T(x_1) + T(x_2)$ work?

share|cite|improve this question
In the definitions, $u$ and $v$ are vectors, not scalars. $T$ is a function of the vector $(x_1,x_2)$. – Alex Becker Sep 12 '12 at 4:41
The map $T(x_1,x_2) = (x_1-2x_2,x_1+5x_2)$ is linear. Where do you the problem lies now that you have a square in the $x_1$ term in the first entry? – user38268 Sep 12 '12 at 4:44
The $x_1^2$ tells you it won't be linear, and suggests some numbers to try to show non-linearity. – Gerry Myerson Sep 12 '12 at 4:45
up vote 2 down vote accepted

Try testing $T((x_1,x_2)+(y_1,y_2))$ and see if it equals $T((x_1,x_2))+T((y_1,y_2))$.

share|cite|improve this answer
Okay, I did... T(4+1, 2+2) = T(5, 4) = (17, 25); and T((4, 2)) + T((1, 2)) = (12, 14) + (-3, 8) = (9, 22); proof: (17, 25) != (9, 22)... correct? – Grace C Sep 12 '12 at 5:10
@LearningPython: The example works, but you need to check your calculation of $T((1,2))$. – Brian M. Scott Sep 12 '12 at 5:13
Okay, fixed it. – Grace C Sep 12 '12 at 5:26

$T(4,2)=(12,14)$, $T(2,1)=(2,7)$,

$2T(2,1)\neq T(4,2)$

so T is not linear.

share|cite|improve this answer
I think T(4, 2) would actually be (12, 14), but your point still holds. – Grace C Sep 12 '12 at 4:59
@LearningPython ,it is a typo.thanks – noname1014 Sep 12 '12 at 5:02
Would this proof also be correct? T(4+1, 2+2) = T(5, 4) = (17, 25); and T((4, 2)) + T((1, 2)) = (12, 14) + (-3, 8) = (9, 22); proof: (17, 25) != (9, 22) – Grace C Sep 12 '12 at 5:12
@LearningPython yes it is right. – noname1014 Sep 12 '12 at 5:16
Yay! Thank you. – Grace C Sep 12 '12 at 5:17

The transformation $T$ in fact may be linear. We are not told what the domain of $T$ and the field of scalars is! If $F$ is a field of characteristic 2, then $(a+b)^2=a^2+b^2$ holds for all $a,b\in F$. However, $(c\cdot x)^2=c\cdot x^2$ with $x\ne0$ implies $c=0\lor c=1$ in any field. Hence we see that $T\colon F\times F\to F\times F$ is not $F$-linear, but atleast it is $\mathbb F_2$-linear.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.